Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $k = \dfrac{q^2 - q - 90}{-4q^2 - 20q} \div \dfrac{q - 10}{q + 5} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{q^2 - q - 90}{-4q^2 - 20q} \times \dfrac{q + 5}{q - 10} $ First factor the quadratic. $k = \dfrac{(q - 10)(q + 9)}{-4q^2 - 20q} \times \dfrac{q + 5}{q - 10} $ Then factor out any other terms. $k = \dfrac{(q - 10)(q + 9)}{-4q(q + 5)} \times \dfrac{q + 5}{q - 10} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (q - 10)(q + 9) \times (q + 5) } { -4q(q + 5) \times (q - 10) } $ $k = \dfrac{ (q - 10)(q + 9)(q + 5)}{ -4q(q + 5)(q - 10)} $ Notice that $(q + 5)$ and $(q - 10)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ \cancel{(q - 10)}(q + 9)(q + 5)}{ -4q(q + 5)\cancel{(q - 10)}} $ We are dividing by $q - 10$ , so $q - 10 \neq 0$ Therefore, $q \neq 10$ $k = \dfrac{ \cancel{(q - 10)}(q + 9)\cancel{(q + 5)}}{ -4q\cancel{(q + 5)}\cancel{(q - 10)}} $ We are dividing by $q + 5$ , so $q + 5 \neq 0$ Therefore, $q \neq -5$ $k = \dfrac{q + 9}{-4q} $ $k = \dfrac{-(q + 9)}{4q} ; \space q \neq 10 ; \space q \neq -5 $